I already proved that $\gcd(a,b) \leq \gcd(a+b,a-b) \leq 2\gcd(a,b)$. I need to prove that we have $\gcd(a+b,a-b)=\gcd(a,b)$ or $\gcd(a+b,a-b)=2\gcd(a,b)$.
I know this is true for when $\gcd(a,b)=1$, since when $a,b$ are of different parity, $\gcd(a+b,a-b)=1$ and when they both are odd, $\gcd(a+b,a-b)=2$. However, I am struggling to prove that this is true for any pair of integers $a,b$ where $\gcd(a,b)>1$.
Based on numerical results (obtained on Mathematica) I think if WLOG $a$ is odd, then $\gcd(a+b,a-b)=2\gcd(a,b)$ when $b$ is odd and $\gcd(a+b,a-b)=\gcd(a,b)$ when $b$ is even. My main trouble comes from when both $a$ and $b$ are even, since I can't seem to easily generalize the results based off what Mathematica gives me as an output.
Any suggestions/ideas on how to approach this problem would be greatly appreciated.